What is the vertices 4x^2+y^2-48x-4y+48=0 ?

The vertices 4x^2+y^2-48x-4y+48=0 is (6,12),(6,-8)

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vertices 4x^2+y^2-48x-4y+48=0

vertices

4 x^{2} + y^{2} - 48 x - 4 y + 48 = 0

(6, 12), (6, - 8)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

4 x^{2} + y^{2} - 48 x - 4 y + 48 = 0 :

Ellipse with center

(h, k) = (6, 2), b = 10, a = 5

(6, 2 + 10), (6, 2 - 10)

Refine

(6, 12), (6, - 8)

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{5} = 1

7 x^{2} + 9 y^{2} - 63 = 0

\frac{5}{81} x^{2} + \frac{y ^{2}}{9} = 1

x^{2} + y^{2} = 1

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

What is the vertices 4x^2+y^2-48x-4y+48=0 ?
The vertices 4x^2+y^2-48x-4y+48=0 is (6,12),(6,-8)